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Parkland College
A with Honors Projects
Honors Program
2014
Calculating P-Values
Isela Guerra
Parkland College
Recommended Citation
Guerra, Isela, "Calculating P-Values" (2014). A with Honors Projects. 107.
http://spark.parkland.edu/ah/107
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Isela Guerra
Instructor: L. Smith
Calculating P- values
Part 1:
Ex: Employees in an accounting firm claim that the mean salary of the firms accountants is less than that
of its competitor’s which is $45,000. (Assume that the standard deviation of the firm’s accountants is
known to be $5,200.) A random sample of 30 of the firm’s accountants has a mean salary of $43,500.
Test the employees claim at the .01 level of significance.
Step 1: H0: µ = 45000
H1: µ < 45000
Step 2: α = .01
Using Z-Test
Step 3:
Figure 1
[Because this was a test about the mean, the population standard deviation is known and n≥30, the Ztest was selected. We are attempting to test the claim that the mean salary is less than $45,000. Our
null hypothesis is then equal to $45,000. The argument is that the mean is less than $45,000 this
becomes our alternative hypothesis.
The level of significance is already given, α = .01
Because there is no data set we use the Z- Test, to get there we click “STATS” and scroll over to "TEST"
then scroll down to "Z-Test". This asks for the null hypothesis in the form of “µ0:” the standard deviation
denoted by “σ”, the mean “x̄”, and “n” the sample size. Now, because we want to know if the
mean salary is less than that of the competing firm we chose “µ: <µ0”. We then click enter on
“calculate” and get our p- value of .0571.]
Using Normal CDF
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Isela Guerra
Instructor: L. Smith
[We want to know what the area to the left of 43,500 is. To do this we use normalcdf and the syntax
within normalcdf is (x̅, x̅µ, µ, σ/ (n) ^1/2). Because we want to know the area to the left of $4, 3500 (µ <
45000) we say, .5-normalcdf(43500,4500,4500,5200/(30)^1/2. Our p-value is .0571]
Using Table and calculating Z value
[To get the z score we use the equation,
, where x̅ is the sample mean and µ is the
population mean, and σ/(n)^1/2 is the population standard deviation divided by the square root of the
sample size. This gave us a Z-score of -1.58 when rounded to three significant figures. To find the
corresponding p-value we go to row labeled Z, scroll down to -1.5 and to the right until we go down to
the row containing .008.]
Step 4: p-value > α
Fail to reject the null hypothesis.
Step 5: There is not significant evidence at the .01 level of significance to conclude that the mean salary
of the accountants is less than that of the competitors.
Part2:
Ex: Account empts survey 150 executives it showed that 44% of them say that “little or no knowledge of
the company” is most common mistake made by candidates during job interviews (based on data from
USA Today). Use a .05 significance level to test claim that less than half of all executives identify that
error as being most common job interviewing error.
Step 1: H0: p = .5 H1: p < .5
Step 2: α = .05
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Isela Guerra
Instructor: L. Smith
Step 3:
Using 1PropZTest
[We use 1-PropZTest because we have to deal with proportions. We are given the proportion we are to
test, p0: .5, we are also given the proportion of those with the given characteristic we are testing for,
which is .44 however to get X we multiply (.44) (150) and get 66. We enter these values into 1-PropZTest
and get a p- value of .0708]
Using Normal CDF
[Because there was no standard deviation given, the normal cdf could not be used until the curve was
standardized. To do this the Z score was calculated by conducting the test statistic, the p with a carrot
above it is the population proportion with given characteristic(.44), and p0 is the population
proportion(.5). The mean is then equal to zero and the standard deviation is equal to 1. Our value
calculated was -1.47 when rounded we then use this value for the normalcdf and get our p-value of
.0708 when rounded.]
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Isela Guerra
Instructor: L. Smith
Using Table and calculating Z value
[Using the value found from the test statistic, the p-value was located on the table. The calculated zvalue is -1.47 when rounded. Go to row labeled Z and stop when -1.4 is hit, proceed to the right when
the row .07 is reached. The p-value is .0708.]
Step 4: p-value > α
Fail to reject the null hypothesis.
Step 5: There is not significant evidence at the .05 level of significance to claim that less than half of all
executives identify that error as being most common job interviewing error.
Part 3:
Ex: Tests in past Statistics classes have had scores with a standard deviation equal to 14.1. A sample of
27 scores from the current classes has a standard deviation of 9.3. Use the 0.01 level of significance to
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Isela Guerra
Instructor: L. Smith
test the claim that this current class is more consistent that past classes. Assume that test scores are
normally distributed.
Step 1: H0: µ = 14.1
H1: µ < 14.1
Step 2: α = .01
Step 3:
Using Test1 SD
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Isela Guerra
Instructor: L. Smith
[This test can be found under programs. Because we were given a summary of the stats we select choice
2 under "The sample info" We press enter and then enter 27 for the sample size and again press enter.
We then move along to the sample standard deviation and enter 9.3 and press enter then select "<" for
less than because we are testing to see if 14.1 is < current classes. We would then like to calculate the pvalue, after clicking enter we then press 2 to select "P-value Method". After, we enter the population
standard deviation of 14.1. The p-value calculated is .00556.]
Using Normal Cdf (X2 Chi square)
[To calculate the p-value using X2(chi Square) cdf, we first calculate the critical value. To do this we use
the formula X2= ((n-1) s2)/σ2. Where (n-1) is the degrees of freedom, s is the sample standard deviation
and σ is the population standard deviation, however when we square this it then becomes the
population variance (σ2). After calculating, we get the critical value of 11.311 when rounded. We then
use this value in X2cdf (0, critical value, degrees of freedom). This then gives us the p- value; we
calculated a p-value of .00556 when rounded to three significant figures.]
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Isela Guerra
Instructor: L. Smith
Using Chi Square table
[When using the table we see degrees of freedom to the top left. We scroll down to 26 degrees of
freedom, and we then see the p-value to the top right of .005.]
Step 4: p-value ≤ α
Reject the null hypothesis
Step 5: There is enough evidence at the .01 level of significance to conclude that this current class is
more consistent that past classes.
Part4
EX: A firm wishes to estimate the economic composition of a community where they are opening a
branch office. They have examined data concerning this and similar communities and have obtained a
hypothetical distribution of income that they feel to be accurate. To check this accuracy, they take a
random sample of 500 families and family incomes. They have hypothesized the following distribution of
income levels:
Category
Income level
Proportion
1
3000+
0.4
2
25000-29999
0.2
3
20000-24999
0.2
4
15000-19999
0.1
5
Under 15000
0.1
The sample yields the following results:
Category
1
2
3
Number
166
97
134
Do these results differ from the hypothesized distribution?
Step 1: H0: p1=0.4, p2=0.2, p3=0.2, p4=0.1, p5=0.1
4
61
5
42
Total
500
H1: The proportions differ from those specified.
Step 2: α=0.05
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Isela Guerra
Instructor: L. Smith
Step 3:
Using X2- Goodness of Fit Test
[To perform this test we place the numbers observed from each of the categories into L1, we then
calculate the proportions which will go into L2. To do this we take the given proportions for each
number and multiply them by the total. For example, (.4) (500) = 200. You can see that 200 is the first
number on the list of L2. We then go to X2 GOF- Test and enter L1 in observed and L2 in expected, we
then scroll down to df (degrees of freedom, found by (n-1)) and enter 4, then enter on calculate and
record our p-value of 2.98X10-4. ]
Using GOF program
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Isela Guerra
Instructor: L. Smith
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Isela Guerra
Instructor: L. Smith
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Isela Guerra
Instructor: L. Smith
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Isela Guerra
Instructor: L. Smith
[To perform the GOF, we go down to programs and enter the information it asks for, however, keep in
mind that when it asks for the null hypothesis we are not expecting them to be all equal, rather equal to
the proportions given, so we enter 2 and then enter each individual proportion to its corresponding
category. At the end of this program we are given a p- value of 2.98X10-4.]
Using Table
[To find the p-value using the table, first we must calculate the chi-square test statistic: X2=SUM
(observed frequency-expected frequency) 2/expected frequency). We take the observed frequency and
subtract it from the proportion we expect multiplied by the total number and square the quantity, we
then divide it by the expected frequency. After calculating the summation we get a chi-square test
statistic value of 21.13. When we take a look at the table we will see that we have a p-value less than
.001 and will not be displayed on the table itself.]
Step 4: p-value ≤ α
Reject the null hypothesis.
Step 5: There is significant evidence at the .05 level of significance to conclude that the proportions are
different.
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